Use a known complex root to recover and factor a cubic

Substitute the given root:

$\begin{aligned}(2+i)^3+a(2+i)^2+2+i+b&=0,\\(2+11i)+a(3+4i)+2+i+b&=0,\\2+11i+3a+4ai+2+i+b&=0,\\(4+3a+b)+(12+4a)i&=0.\end{aligned}$

For efficiency, a graphing calculator in complex mode can evaluate $(2+i)^2=3+4i$ and $(2+i)^3=2+11i$; this follows the source guidance. Comparing real and imaginary parts still supplies the equations $4+3a+b=0$ (1) and $12+4a=0$ (2). From (2), $a=-3$; then (1) gives $b=5$.

Now $f(z)=z^3-3z^2+z+5$ has real coefficients. Since $2+i$ is a root, $2-i$ is also a root. The factors $[z-(2+i)]$ and $[z-(2-i)]$ combine to

$\begin{aligned}&[z-(2+i)][z-(2-i)]\\ & =(z-2)^2-i^2\\ & =z^2-4z+5.\end{aligned}$

Write $f(z)=(z^2-4z+5)F(z)$. Long division starts with quotient term $z$. Multiply the divisor by $z$ and subtract:

$\begin{aligned}z(z^2-4z+5)&=z^3-4z^2+5z,\\(z^3-3z^2+z+5)-(z^3-4z^2+5z)&=z^2-4z+5.\end{aligned}$

The next quotient term is $1$. Subtract $1(z^2-4z+5)$ from the remainder to leave zero, so $F(z)=z+1$. Thus

$\begin{aligned}&z^3-3z^2+z+5\\ & =[z-(2+i)][z-(2-i)][z-(-1)].\end{aligned}$

The other roots are $2-i$ and $-1$. This is one real root and one nonreal conjugate pair; a real cubic otherwise has three real roots, counting multiplicity.

On the calculator illustrated in the notes:

1. Press apps and select PlySmlt2.
2. Select POLYNOMIAL ROOT FINDER.
3. Select Order 3 and a+bi, so nonreal roots appear in algebraic form.
4. Enter the coefficients $1,-3,1,5$ for $z^3-3z^2+z+5=0$.
5. Press graph to solve. The screen lists $x_1=-1$, $x_2=2-i$, $x_3=2+i$.

This verifies the roots obtained by the exact factorisation; retain the algebra when explaining why these are the roots.